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Question
A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings Specific heat of aluminium = 0.91 J g–1 K–1
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Solution 1
Power of the drilling machine, P = 10 kW = 10 × 103 W
Mass of the aluminum block, m = 8.0 kg = 8 × 103 g
Time for which the machine is used, t = 2.5 min = 2.5 × 60 = 150 s
Specific heat of aluminium, c = 0.91 J g–1 K–1
Rise in the temperature of the block after drilling = δT
Total energy of the drilling machine = Pt
= 10 × 103 × 150
= 1.5 × 106 J
It is given that only 50% of the power is useful.
Useful energy, `triangle Q = 50/100 xx 1.5 xx 10^6 = 7.5xx10^5 J`
But `triangle Q = mctriangle T`
`:. triangle T = (triangle Q)/"mc"`
`= (7.5 xx 10^5)/(8xx10^3xx0.91)`
`= 103 ^@C`
Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is 103°C.
Solution 2
Power = 10 kW = 104 W
Mass, m=8.0 kg = 8 x 103 g
Rise in temperature, `triangle T =?`
`Time, t = 2.5 min = 2.5 xx 60 = 150 s`
Specific heat, `C = 0.91 Jg^(-1) K^(-1)`
Total energy = Power x Time = `10^4 xx 150 J`
`=15 xx 10^5 J`
As 50% of energy is lost
∴Thermal energy available
`triangle Q = 1/2 xx 15 xx 10^5 = 7.5 xx 10^5 J`
Since `triangle Q = mctrangle T`
`:.triangle T = triangleQ/mc = (7.5xx20^5)/(8xx10^3xx0.91) = 103^@C`
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